Confused about proof with rational/irrational numbers

Question

I'd like to solve this simple task but I am stuck:

Prove the following statement by proving its contrapositive: if $r$ is irrational, then $r^{1/5}$ is irrational.

I think the contrapositive would be: If $r^{1/5}$ is rational, then $r$ is rational. But I am confused. Doesn't $r^{1/5}$ always turn into the $5$^th root of $r$ which makes the result always irrational?

Answer 1

No, $r^{1/5}$ isn't automatically irrational. Just look at $r = 32$. Then $r^{1/5} = 2$.

And yes, your contrapositive is correctly stated. Your proof should start something like this:

Suppose that $r^{1/5}$ is rational, so $r^{1/5} = \frac{a}{b}$ for some integers $a$ and $b$. But then $r = \cdots $

Answer 2

$$ r\notin\Bbb Q\Rightarrow\sqrt[5]r\notin\Bbb Q\iff\sqrt[5]r\in\Bbb Q\Rightarrow r\in\Bbb Q$$

$$\sqrt[5]r\in\Bbb Q\iff \sqrt[5]r=\frac ab \text{ where } a,b\in\Bbb Z, b\ne 0$$ This implies $$r=\frac{a^5}{b^5}$$ where $a^5$ and $b^5$ satisfy the same conditions that $a$ and $b$, which means that $r$ is rational.